arianna1453
  • arianna1453
I will fan!! I need help with some calculus questions.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
arianna1453
  • arianna1453
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TrojanPoem
  • TrojanPoem
To find y' you have to derive the function with respect to x \[x = \sin(x+y)\] \[1 = \cos(x+y) . (1 + y')\] \[\frac{ 1 }{ \cos(x+y)} = 1 + y'\] \[\frac{ 1 }{ \cos(x+y)} - 1 = y'\] \[y' = \frac{ 1 - \cos(x+y) }{ \cos(x+y) }\]
arianna1453
  • arianna1453
Thank you. I got that right. Could you check this one?
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arianna1453
  • arianna1453
@TrojanPoem
TrojanPoem
  • TrojanPoem
The first, second are wrong: ddx[f(x)g(x)]=f′(x)g(x)+g′(x)f(x) The third: 0.5 (f(x))^-0.5 * f'(x) = f'(x) / 2sqrt(fx) Maybe you noticed something, I didn't. why do you think the third is wrong ?
arianna1453
  • arianna1453
I knew it was either the third or none of them. The third one is confusing to me.
TrojanPoem
  • TrojanPoem
The second derivative of absolute is :\[\frac{ u.u' }{ |u| }\] In the third: \[\frac{ d }{ dx } \sqrt{f(x)} = \frac{ d }{ dx } f(x)^{0.5} = 0.5 . f(x)^{0.5 - 1} * f'(x)\] \[0.5. f(x)^{-0.5} * f'(x) = \frac{ f'(x) }{ 2 \sqrt(fx) }\]
arianna1453
  • arianna1453
Okay, so the third one is the true statement.
TrojanPoem
  • TrojanPoem
I believe so.
arianna1453
  • arianna1453
Thank you. it was correct! (:
TrojanPoem
  • TrojanPoem
Any time.

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